Question

A class consists of a number of boys whose ages are AP. The common difference begin 4 months .if the youngest boys is 8 years and if the sum of ages 168 years then the number of boys is

Answer 1

Heyy Buddy

Here's your Answer

Given :-

Youngest boy's age = a = 8.

d = 4 month = 1/3 yrs.

Sn = 168

Now,

Sn = n/2 [ 2a + (n-1) d ]

=> 168 = n/2 [ 16 + (n-1) 1/3]

=> 336 = n [ 16 + n/3 - 1/3]

=> 336 = n ( 48 + n - 1)/3

=> 336 × 3 = n ( n + 47)

=> 1008 = n^2 + 47n

=> n^2 + 47n - 1008 = 0

=> n^2 + (63 - 16)n - 1008 = 0

=> n( n + 63) - 16 ( n + 63) = 0

=> (n + 63) ( n - 16)

=> n = -63 ( Neglect) and n = 16.

Therefore,

There are total 16 boys in the class.
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